# Trigonometric form of complex numbers

Let $M (x,y)$ be the point in the complex plane which is join to the complex number $z = x + yi$.

The position of the point $M$, and thus of the complex number $z$, we can determine by using numbers $r$ and $\varphi$ where $r = |z|$ (the distance from the point $M$ to the origin) and $\varphi \in \left [0, 2\pi \right \rangle$ is an angle which the segment line $\overline{OM}$ closing with the positive part of the real axis.

Then

$$\cos \varphi = \frac{x}{r} \Rightarrow x = r \cos \varphi$$

and

$$\sin \varphi = \frac{y}{r} \Rightarrow y = r \sin \varphi$$

is valid.

Substituting in the expression $z=x + yi$ we obtain trigonometric form of  the complex number:

$$z= r (\cos \varphi + i \sin \varphi).$$

If  a complex number is given in the algebraic form $z = x+yi$, then $r$ and $\varphi$ we determine from equations:

$$r = \sqrt{x^2 + y^2}$$

$$tg \varphi = \frac{y}{x} , x \neq 0.$$

The last equation gives two solutions for an angle $\varphi \in \left [0, 2\pi \right \rangle$. We choose the angle depend on in which quadrant the number $z$ is located. The angle $\varphi$ is called an argument of the complex number $z= x + yi$ and is denoted as $arg(z)$.

Multiplication and division of complex numbers in trigonometric form

If $z_1 = r_1 ( \cos \varphi_1 + i \sin \varphi_1)$ and $z_2 = r_2 ( \cos \varphi_2 + i \sin \varphi_2)$, $z_2 \neq 0$, then their product is define as

$$z_1 \cdot z_2 = r_1 r_2 ( \cos (\varphi_1 + \varphi_2) + i \sin (\varphi_1 + \varphi_2),$$

and their quotient as

$$\frac{z_1}{z_2} = \frac{r_1}{r_2} ( \cos (\varphi_1 – \varphi_2) + i \sin (\varphi_1 – \varphi_2).$$

Example 1. Write in the trigonometric form complex numbers $z$ and $\overline{z}$ if $z = \frac{1}{2} – \frac{\sqrt{3}}{2}i$.

Solution:

We need to determine numbers $r$ and $\varphi$.

$$r= |z|$$

$$|z| =|\frac{1}{2} – \frac{\sqrt{3}}{2}i|$$

$$= \sqrt{\left( \frac{1}{2} \right) ^2 + \left( -\frac{\sqrt{3}}{2} \right)^2 }$$

$$= \sqrt{\frac{1}{4} + \frac{3}{4}}$$

$$= \sqrt{1}$$

$$= 1.$$

$$tg \varphi = \frac{-\frac{\sqrt{3}}{2}}{\frac{1}{2}} = -\sqrt{3},$$

that is $\varphi = \frac{5\pi}{3}$ or $\varphi = \frac{8\pi}{3}$. Since the number $z = \frac{1}{2} – \frac{\sqrt{3}}{2}i$ is located in the fourth quadrant, it follows that $\varphi = \frac{5\pi}{3}$.

The complex number $z= \frac{1}{2} – \frac{\sqrt{3}}{2}$ has the trigonometric form:

$$z = \cos \frac{5 \pi}{3} + i \sin \frac{5\pi}{3}.$$

The complex conjugate numbers have the same modulus, therefore, for $\overline{z} = \frac{1}{2} + \frac{\sqrt{3}}{2}$ , $r =1$. $\overline{z}$ is located in the first quadrant, so we have:

$$tg \varphi = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3} \Rightarrow \varphi = \frac{\pi}{3}.$$

Finally, the complex number $\overline{z} = \frac{1}{2} + \frac{\sqrt{3}}{2}$ has the following trigonometric form:

$$\overline{z} = \cos \frac{\pi}{3} + i \sin \frac{\pi}{3}.$$

Example 2. Write in the trigonometric form the complex number $z$ where:

$$z = – \cos \frac{\pi}{5} + i \sin \frac{\pi}{5}.$$

Solution:

The function cosine is negative in the second and third quadrant, and sine is positive in the first and second quadrant. This means that the given complex number $z$ is located in the second quadrant.

Now we have:

$$tg \varphi = \frac{- \cos \frac{\pi}{5}}{\sin \frac{\pi}{5}} = – tg \frac{\pi}{5}.$$

That is, $\varphi = – \frac{\pi}{5}$ or $\varphi = \frac{4\pi}{5}$.

We know that the complex number $z$ is located in the second quadrant, which means that $\varphi = \frac{4\pi}{5}$. The radius of the same complex number is equal to $1$.

Now we can write the given complex number $z$ in the trigonometric form:

$$z = \cos \frac{4 \pi}{5} + i \sin \frac{4 \pi}{3}.$$

De Moivre’s formula

For every complex number $z = r (\cos \varphi + i \sin \varphi)$ and $n \in \mathbb{N}$

$$z^n = r^n ( \cos n \varphi + i \sin n \varphi)$$

is valid.

The formula above is called the De Moivre’s formula.

Example 3. Calculate

$$\frac{(-1 + i)^{20}}{(\sqrt{2} – i \sqrt{2})^{10}}.$$

Solution:

Let  $z_1 = -1 + i$ and $z_2 = \sqrt{2} – i \sqrt{2}$.

$z_1$ has the following trigonometric form:

$$r_1 = \sqrt{(-1)^2 + 1^2} = \sqrt{2}$$

$$tg \varphi_1 = -1 \Rightarrow \varphi_1 = \frac{3\pi}{4},$$

because $z_1$ is in the second quadrant. Therefore

$$z_1 = \sqrt{2} ( \cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4}).$$

Now, by using the De Moivre’s formula, we have:

$$z_1^{20} = (-1 + i)^{20}$$

$$=(\sqrt{2})^{20} \left ( \cos \left(20\cdot \frac{3\pi}{4}\right) + i \sin \left(20\cdot \frac{3\pi}{4}\right)\right)$$

$$= 1024 ( \cos (15\pi) + i \sin (15\pi)$$

$$= 1024 (\cos \pi + i \sin \pi)$$

For $z_2$ we have:

$$r_2 = \sqrt{(\sqrt{2})^2 + (\sqrt{2})^2} = \sqrt{2+2} = \sqrt{4} = 2$$

$$tg \varphi_2 = -1 \Rightarrow \varphi_2 = \frac{7\pi}{4},$$

because the number $z_2$ is located in the fourth quadrant. It follows that the trigonometric form of the complex number $z_2$ is:

$$z_2 = 2 ( \cos \frac{7\pi}{4} + i \sin \frac{7\pi}{4}).$$

By using the De Moivre’s formula we have:

$$z_2^{10} = (\sqrt{2} – i \sqrt{2})^{10}=$$

$$= 2^{10} \left ( \cos \left(10\cdot \frac{7\pi}{4}\right) + i \sin \left(10\cdot \frac{7\pi}{4}\right)\right)$$

$$= 1024 \left ( \cos \left(\frac{35 \pi}{2}\right) + i \sin \left(\frac{35 \pi}{2}\right) \right)$$

$$= 1024 \left ( \cos \left(\frac{\pi}{2}\right) + i \sin \left(\frac{\pi}{2}\right) \right)$$

The rest is to calculate the quotient:

$$\frac{z_1^{20}}{z_2^{10}}$$

$$= \frac{(-1 + i)^{20}}{(\sqrt{2} – i \sqrt{2})^{10}}$$

$$=\frac{1024}{1024 } \cdot \left ( \cos \frac{\pi}{2} + i \sin \frac{\pi}{2} \right)$$

$$= 1 \cdot \left ( \cos \frac{\pi}{2} + i \sin \frac{\pi}{2} \right)$$

$$= i$$

$n$th root of complex numbers

Every complex number $z= r (\cos \varphi + i \sin \varphi)$ has exactly  $n$ different values of $n$-th root given by the formula:

$$\sqrt[n]{z} = \sqrt[n]{r} \left ( \cos \frac{\varphi + 2 k \pi}{n} + i \sin \frac{\varphi + 2 k \pi}{n} \right),$$

where $k \in \{0, 1, 2, …, n-1\}$ and $n \in \mathbb{N}$.

Example 4.  Determine $\sqrt[6]{1}$.

Solution:

The number $z=1 + 0\cdot i$ has the following trigonometric form:

$$r = \sqrt{1^2 + 0^2} = \sqrt{1} = 1$$

$$tg \varphi = 0 \Rightarrow \varphi = 0$$

$$\Rightarrow z = 1 (\cos 0 + i \sin 0).$$

Now we have, by using the formula for the $n$th root of a complex number for $n=6$:

$$\sqrt[6]{1} = \sqrt[6]{1} \left ( \cos \frac{0 + 2k\pi}{5} + i \sin \frac{0 + 2 k \pi}{5} \right).$$

For $k=0$:

$$z_0 = 1 (\cos 0 + i \sin 0) = 1.$$

For $k = 1$:

$$z_1 = 1 \left( \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} \right) = \frac{1}{2} + i \frac{\sqrt{3}}{2}.$$

For $k=2$:

$$z_2 = 1 \left ( \cos \frac{2 \pi}{3} + i \sin \frac{2 \pi}{3} \right) = – \frac{1}{2} + i \frac{\sqrt{3}}{2}.$$

For $k=3$:

$$z_3 = 1 (\cos \pi + i \sin \pi) = -1.$$

For $k=4$:

$$z_4 = 1 \left ( \cos \frac{4\pi}{3} + i \sin \frac{4\pi}{3} \right ) = – \frac{1}{2} – i \frac{\sqrt{3}}{2}.$$

For $k=5$:

$$z_5 = 1 \left ( \cos \frac{5\pi}{3} + i \sin \frac{5\pi}{3} \right) = \frac{1}{2} – i \frac{\sqrt{3}}{2}.$$

As we can observe, obtained complex numbers are arranged symmetrical. They form the vertices of a regular hexagon.

Similar, holds true in general. The formula for the $n$th root of complex numbers has a geometric interpretation. All $n$th roots of complex numbers have the same modulus $\sqrt[n]{r}$, which means that they are equally distant from the origin. Therefore, they all lie on the circle with center at the origin and radius $\sqrt[n]{r}$.

Furthermore, arguments of every two consecutive $n$th roots are differ for $\frac{2 \pi}{n}$, which means that these $n$th roots are arranged so that they form the vertices of a regular polygon.